# Prove some identities! (Round two)

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• Jan 13th 2011, 01:46 AM
Drexel28
Quote:

Originally Posted by Random Variable
Sorry. I have another one.

$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{(1+e^{x})^{2}} \ dx \ , \ a>0$

Thanks!

Spoiler:

Let

$\displaystyle I(a)=\int_0^{\infty}\frac{x^a}{\left(1+e^x\right)^ 2}\text{ }dx$

Note then that

$\displaystyle I(a)=\int_0^{\infty}\frac{e^{-2x}x^a}{\left(1+e^{-x}\right)^2}\text{ }dx$

Noticing though (this isn't in full rigor, but it can be justified via limits) that for our bounds we have that

$\displaystyle \frac{1}{(1+e^{-x})^2}=\sum_{n=1}^{\infty} ne^{-(n-1)x}(-1)^n$

We may conclude that

\displaystyle \begin{aligned}I(a) &= \int_0^{\infty}\frac{e^{-2x}x^a}{(1+e^{-x})^2}\text{ }dx\\ &= \int_0^{\infty}e^{-2x}x^a\sum_{n=1}^{\infty}ne^{-(n-1)x}(-1)^n\text{ }dx\\ &=\sum_{n=1}n(-1)^n\int_0^{\infty}x^a e^{-(n+1)x}\text{ }dx\\ &= \Gamma(a+1)\sum_{n=1}^{\infty}\frac{n(-1)^n}{(n+1)^{a+1}}\\ &=\Gamma(a+1)\left\{\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)^a}-\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)^{a+1}}\right\}\\ &= \Gamma(a+1)\left(\left(1-2^{-a}\right)\zeta(a+1)-\left(1-2^{-(a+1)}\right)\zeta(a)\right)\end{aligned}

• Jan 13th 2011, 11:01 AM
Random Variable
My solution only involves $\zeta(a)$ and $\Gamma(a+1)$. Maybe your solution can be simplified.

EDIT: I found an error in my solution. Your solution is most likely correct.
• Jan 13th 2011, 06:06 PM
Random Variable
I would never have thought to break up the summation like that.
• Jan 14th 2011, 07:50 AM
PaulRS
Here's a nice series:

Compute: $\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}$
• Jan 15th 2011, 12:08 PM
Random Variable
Quote:

Originally Posted by PaulRS
Here's a nice series:

Compute: $\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}$

Perhaps a hint?
• Jan 15th 2011, 12:16 PM
chiph588@
Quote:

Originally Posted by PaulRS
Here's a nice series:

Compute: $\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor}$

This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

$\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}$

$\displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots$

$\displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma$

Edit: Wait... I'm not confident in this solution all of the sudden..
• Jan 15th 2011, 10:20 PM
Drexel28
Quote:

Originally Posted by chiph588@
This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

$\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}$

$\displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots$

$\displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma$

Edit: Wait... I'm not confident in this solution all of the sudden..

I'm not quite sure how the last step (technically second to last, where you conclude that the sum but with $\lfloor \lg(k)\rfloor$ is replaced with $\lfloor\lg(k)\rfloor-n$ tends to zero) is achieved in your proof? Your answer is correct though. I'll leave it for someone else to do, unless no one else wants to?
• Jan 15th 2011, 11:31 PM
chiph588@
Quote:

Originally Posted by Drexel28
I'm not quite sure how the last step (technically second to last, where you conclude that the sum but with $\lfloor \lg(k)\rfloor$ is replaced with $\lfloor\lg(k)\rfloor-n$ tends to zero) is achieved in your proof? Your answer is correct though. I'll leave it for someone else to do, unless no one else wants to?

Yeah, that 'proof' is complete garbage...
• Jan 16th 2011, 04:11 AM
PaulRS
Quote:

Originally Posted by chiph588@
This proof is invalid because the series is conditionally convergent, but perhaps it can be extended by introducing an error term of some sort.

$\displaystyle\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot \lfloor\log_2(k) \rfloor} = 1 - \log2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-1)}$

$\displaystyle = 1+\tfrac12 - \log2^2 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-2)} = 1+\tfrac12+\tfrac13+\tfrac14 - \log2^3 +\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-3)} = \ldots$

$\displaystyle = H_{2^n} - \log2^n + \sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot (\lfloor\log_2(k) \rfloor-n)} = \lim_{n\to\infty}\left(H_{2^n-\log2^n}\right) = \gamma$

Edit: Wait... I'm not confident in this solution all of the sudden..

If you read carefully, the first line implies: $\log(2) = \frac{1}{2}$ ... problem with the signs and the indices of the sums.

Hint:
Spoiler:
try to exploit that $\lfloor \log_2(n) \rfloor$ is constant in certain intervals.
• Jan 16th 2011, 01:54 PM
chiph588@
Quote:

Originally Posted by PaulRS
try to exploit that $\lfloor \log_2(n) \rfloor$ is constant in certain intervals.

That's what I tried to do above... I just did a very bad job at it haha. Let me try again.

Define $\displaystyle s_n = \sum_{k=1}^{2^n}\frac{(-1)^{k-1}}k$ and $r_n = \log2-s_n$.

Observe $\displaystyle \lim_{n\to\infty} nr_n = 0 \implies \gamma = \lim_{n\to\infty} (H_{2^n}-n\log2) = \lim_{n\to\infty} (H_{2^n}-ns_n)$.

Now note $\displaystyle H_{2^n}-ns_n - H_2-s_1 = \sum_{k=1}^{n-1} (H_{2^{k+1}}-(k+1)s_{k+1}-H_{2^k}+ks_k)$ implies $\displaystyle \gamma = 1-\sum_{n=1}^\infty n(s_{n+1}-s_n) = 1-\sum_{n=1}^\infty \sum_{k=2^{n-1}+1}^{2^n} \frac n{2k(2k+1)}$

------

Going along the hint we get $\displaystyle \sum_{k=1}^{\infty}{\frac{(-1)^k}{k}\cdot \lfloor\log_2(k)\rfloor = \sum_{n=1}^\infty n\sum_{k=2^n}^{2^{n+1}-1} \frac{(-1)^k}{k} = \sum_{n=1}^\infty n\sum_{k=2^{n-1}}^{2^n-1} \frac1{2k(2k+1)}$

$\displaystyle = 1-\sum_{n=1}^\infty \sum_{k=2^{n-1}+1}^{2^n} \frac n{2k(2k+1)} = \gamma$
• Jan 16th 2011, 03:48 PM
chiph588@
Compute $\displaystyle \int_0^1 \frac{1-e^{-t}}t dt -\int_1^\infty \frac{e^{-t}}t dt$
• Jan 16th 2011, 09:00 PM
Drexel28
Quote:

Originally Posted by chiph588@
Compute $\displaystyle \int_0^1 \frac{1-e^{-t}}t dt -\int_1^\infty \frac{e^{-t}}t dt$

Solution:

Spoiler:

Recall the common fact that $\Gamma'(1)=-\gamma$. Note though that

$\displaystyle \Gamma'(x)=\frac{d}{dx}\int_0^{\infty}t^{x-1}e^{-t}\text{ }dt=\int_0^{\infty}\frac{\partial }{\partial x}t^{x-1}e^{-t}\text{ }dt=\int_0^{\infty}t^{x-1}\log(t)e^{-t}\text{ }dt$

and thus

$\displaystyle -\gamma=\int_0^{\infty}\log(t)e^{-t}\text{ }dt=\int_0^1 \log(t)e^{-t}\text{ }dt+\int_1^{\infty}\log(t)e^{-t}\text{ }dt$

integrate by parts in each to get that

$\displaystyle -\gamma=\left((-e^{-t}+1)\log(t)\right)\bigg\mid_0^1-\int_0^1\frac{e^{-t}+1}{t}\text{ }dt+\left(-e^{-t}\log(t)\right)\bigg\mid_1^{\infty}-\int_1^{\infty}\frac{e^{-t}}{t}\text{ }dt$

Note though that both the non-integral evaluations are zero and so we may conclude that

$\displaystyle -\gamma=\int_0^1\frac{e^{-t}-1}{t}\text{ }dt+\int_1^{\infty}\frac{e^{-t}}{t}\text{ }dt$

Multiplying both sides by $-1$ gives the correct result.
• Jan 16th 2011, 09:04 PM
Drexel28
What about $\displaystyle \int_0^{\infty}e^{-x^2}\log(x)\text{ }dx$?
• Jan 16th 2011, 09:30 PM
chiph588@
Quote:

Originally Posted by Drexel28
What about $\displaystyle \int_0^{\infty}e^{-x^2}\log(x)\text{ }dx$?

Spoiler:

Consider $\displaystyle I(a) = \int_0^\infty x^ae^{-x^2} dx$, we're after $\displaystyle I'(0)$.

Let $\displaystyle x=\sqrt{t} \implies dx = \frac{dt}{2\sqrt{t}}$, so $\displaystyle I(a) = \frac12\int_0^\infty t^{(a-1)/2}e^{-t} dt = \frac12\Gamma\left(\frac{a+1}2\right)$.

Thus $\displaystyle I'(a) = \frac14\Gamma'\left(\frac{a+1}2\right)$.

So $\displaystyle I'(0) = \frac14\Gamma'\left(\frac12\right) =-\frac14\sqrt{\pi}(\gamma+2\log2)$. (the last equality is not trivial (Tongueout))

• Jan 16th 2011, 09:33 PM
Drexel28
Quote:

Originally Posted by chiph588@
[spoiler]

(the last equality is not trivial (Tongueout))

So then I suppose it will be impressive when you prove it, right! :)
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